It is not quite that simple. Power drop is not really a factor.
The operating range of the radio, from specifications, is 11 volts minimum to 15 volts maximum. Thus with a 4 volt operating range, a 1 volt drop represents a 25% change. Within this range of 11 to 15 volts the radio will attempt to deliver 100 watts by demanding the required amount of current from the power supply. Thus the three resistances in series example, as I stated earlier, with the middle one {radio} being a variable value.
Therefore the resistance of the DC power cables will then cause more voltage drop to occur as the current demand increases. This has been noted to cause the transmitter IMD {Inter-Modulation Distortion} to significantly increase. Much has been written here regarding attaining lowest IMD while operating at near maximum rated voltage. A word of caution, NEVER allow the power supply to exceed 15 volts. It is very important to maintain minimum voltage drop with the power cables and anything else in the path. Those power distribution strips are horrible in this regard and should NEVER be used to power the radio. The radio should always be connected direct to the power supply terminals. No exceptions.
For this and other reasons, I have my power supply set to 14.8 volts at no load. In transmit with 100 watts output and current of 17.5 amps, the voltage indicated by the radio is 14.5 volts or 0.3 volt drop {14.8 - 14.5 = 0.3}. The calculated resistance of my power cables is then E / I = R or 0.5 / 17.5 = 0.017 ohms.
In conclusion, it is NOT power output that is of concern. 73 Bob, K4TAX On 12/28/2019 1:49 PM, [email protected] wrote:
At what point does a drop in input power become significant? Does a 1.0v loss translate to 10 times less output power than a 0.1v loss? -----Original Message----- From: [email protected] <[email protected]> On Behalf Of Bob McGraw K4TAX Sent: Saturday, December 28, 2019 1:07 PM To: [email protected] Subject: Re: [Elecraft] K3 - Power cable voltage drop Actually 1 ohm of total resistance from a 14V source and a 20A load would be more than excessive. More like a 0.01 ohm is 0.20 volt drop for 20 amps. I x R = E View it as 3 series resistors across a 14 V source. R-1 is the DC POS lead resistance, the radio which is a variable resistance as R-2, and R-3 which is the DC NEG lead resistance. Therefore, the voltage at the radio is equal to the IR drop being the sum of R-1 and R-3. 73 Bob, K4TAX
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